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xenn [34]
2 years ago
10

Object A attracts object B with a gravitational force of 5 newtons from a given distance. If the distance between the two object

s is reduced in
half, what will be the changed force of attraction between them?
Physics
2 answers:
givi [52]2 years ago
5 0

Answer:

Four times higher

Explanation:

F- G (m1 x m2)/ r^2

if r 1  = 2 and r 2 = 1       therefore  F = G (m1 x m2)/  1^2 is 4 times higher than

                                    2^2 since G and m1  and  m2 remained the same

Pepsi [2]2 years ago
5 0

answer is D. 20 newtons for plato, got 100 on the test :)

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Meme on newtons law of motion (should me made ur self not from any searh engine)
k0ka [10]
He will be a pilot and he will fly the plane over bridges fewwww
8 0
1 year ago
Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass g
monitta

Answer:

55.56kg

Explanation:

Given:

F= 52N

a=0.936m/s²

Applyinc Newton's second law, that states: force is equal to mass times acceleration.

F = ma

m=F/a =>52 / 0.936

m=55.56kg

5 0
3 years ago
Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the
Svet_ta [14]

Answer:

b)  x_{cm} = 4.88 cm , c) x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃) and d)

x_{cm}’= 1.88 cm

Explanation:

The definition of mass center is

    x_{cm} = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

    D₂ = d₁ + d₂

    D₂ = 1.1 + 1.9

    D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

    D₃ = D₂ + d₂

    D₃ = 3.0 + 3.9

    D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

    M = m₁ + m₂ + m₃

    M = 24 + 12 + 56

    M = 92 g

    x_{cm} = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

    x_{cm} = 1/92 (448.8)

    x_{cm} = 4,878 cm

    x_{cm} = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at   d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

    x_{cm} ’= 1/M  (m₁ d₁ + m₂ d₂ + m₃ d₃)

   

   x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

    x_{cm}’= 1/92 ((24 (-1.9) +56 3.9)

    x_{cm}’= 1/92 (172.8)

    x_{cm}’= 1.88 cm

This distance is to the right of the central marble

3 0
3 years ago
What’s the answer? Will give brainliest
Stella [2.4K]

Answer:

The answer to your question is: letter D

Explanation:

Fussion: means combine 2 light nucleus to form heavier nucleus.

Fission: means splitting heavy nucleus to form lighter nucleus.

In the figure we can see two light nucleus that combine to form heavier nucleus, so the answer is nuclear fussion.

5 0
3 years ago
Read 2 more answers
A satellite is circling the moon (radius 1700 km) close to the surface at a speed v. A projectile is launched vertically up from
Elina [12.6K]

Answer:

 Rmax = 3.4 10⁶ m

Explanation:

For this exercise we will use the concept of energy

Initial. On the surface of the luma

    Em₀ = K + U

    Em₀ = ½ m v² - G m M / R_moon

Final. At the furthest point

    Emf = U

    Emf = - g m M / R_max

    Em₀ = Emf

    ½ m v² - G m M / R_moon = - G m M / R_max

    ½ v² + G M (-1 / R_moon + 1 / R_max) = 0           (1)

Let's use the fact that tells us that the speed of the rocket is equal to the speed of a satellite that rotates around the moon near the surface, let's use Newton's second law

        F = m a

Acceleration is centripetal

       a = v² / r

        r= R_moon

       G m M / R_moon² = m v² / R_mon

        G M / R_moon = v²

We substitute in 1

     ½ G M / R_moon + G M (1 / R_max - 1 / R_moon) = 0

    1 / R_max = 1 / R_moon (1- ½)

     R_max = R_moon 2

     Rmax = 2 1700 103

     Rmax = 3.4 10⁶ m

5 0
2 years ago
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